Two point charges, X and Y, are held 0,03 m apart, as shown in the diagram below - NSC Physical Sciences - Question 7 - 2023 - Paper 1

The electric field lines will emanate from the positive charge and move towards the negative charge, demonstrating the attraction between them. The pattern will show lines directed from charge Y towards charge X, while charge X will have field lines curling inward, indicating the direction of force experienced by a positive test charge.

Using Coulomb's law, the formula for electrostatic force is given by:

$F = k \frac{|Q_1 Q_2|}{r^2}$

Where:

• $k = 8.99 × 10^9 \frac{N m^2}{C^2}$
• $Q_1 = -7.2 × 10^{-9} C$ (charge of X)
• $Q_2 = 7.2 × 10^{-9} C$ (charge of Y)
• $r = 0.03 m$

Substituting these values, we have:

$F = 8.99 × 10^9 \frac{(7.2 × 10^{-9})(7.2 × 10^{-9})}{(0.03)^2}$ $F = 5.18 × 10^{-4} N$

The vector diagram should illustrate two arrows: one directing towards charge X from charge Y, and another arrow emanating from charge Z towards charge X. These arrows should signify the respective electric fields due to the charges Y and Z at point X.

To find charge Z, we begin by calculating the resultant electric field at point X due to charges X and Y.

The net electric field at point X is given as:

$E_{net} = E_Y + E_Z$

Given that $E_{net} = 4.91 × 10^{5} N/C$ and knowing that:

$E_Y = k \frac{Q_Y}{r^2}$

We can rearrange to find:

$Q_Z = \frac{E_{net} * r^2}{k}$

Substituting the values: $r = 0.01 m$, and solving will yield:

$Q_Z = 6.25 × 10^{-9} C.$